Let $P$ be a point outside of circle $O.$ A segment is drawn from $P$ such that it is tangent to circle $O$ at point $T.$ Meanwhile, a secant from $P$ intersects $O$ at $A$ and $B,$ such that $PA < PB.$ If $PA = 3$ and $PT = AB - PA,$ then what is $PB$?
Solution: First of all, we see that $PB = PA + AB = 3 + AB.$ By Power of a Point, we know that $(PA)(PB) = (PT)^2,$ so we have $3(PB) = (AB - 3)^2.$

[asy]
unitsize(2 cm);

pair A, B, O, P, T;

T = dir(70);
P = T + dir(-20);
B = dir(150);
O = (0,0);
A = intersectionpoint(P--interp(P,B,0.9),Circle(O,1));

draw(Circle(O,1));
draw(T--P--B);

label("$A$", A, SW);
label("$B$", B, NW);
dot("$O$", O, S);
label("$P$", P, E);
label("$T$", T, NE);
[/asy]

Let us define $x$ such that $x = PB = 3 + AB,$ then $AB = x - 3.$ Substituting, we now have $3x = (x - 6)^2.$

Then, we see that $3x = x^2 - 12x + 36,$ so $x^2 - 15x + 36 = 0.$ Factoring, we have $(x - 3)(x - 12) = 0$ so $x = 3$ or $x = 12,$ but we are given that $PA < PB,$ so $x > 3.$ That means our only answer for $x,$ hence $PB,$ is $\boxed{12}.$